Entry 122

Given Gauss' constant \(G\), the elliptic integral singular value \(K(k_r)\), $$\begin{align}G &=\frac{\sqrt2}{\pi} K(k_1) = \frac2{\pi}\int_0^{\pi/2}\frac{d\,\theta}{\sqrt{1+\sin^2\theta}}\\ &= \tfrac1{\sqrt2}\,_2F_1\big(\tfrac12,\tfrac12;1;\tfrac12\big) = \,_2F_1\big(\tfrac12,\tfrac12;1;-1\big)\\ &=(2\pi)^{-3/2}\,\Gamma^2\big(\tfrac14\big) = 0.834626\dots\end{align}$$ and modular lambda function \(\lambda(\tau)\) calculated by Mathematica as ModularLambda[tau]. The two hypergeometrics can be expressed as $$\tfrac1{\sqrt2}\,_2F_1\big(\tfrac12,\tfrac12;1;\lambda(i)\big) = \,_2F_1\big(\tfrac12,\tfrac12;1;\lambda(1+i)\big) = G$$ while more complicated ones are $$\begin{align}\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/1\big)\Big) &= 2^{1/4}G\\ \,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/3\big)\Big) &= 6^{1/4}\left((1+\sqrt{3})^2-\sqrt{(2+\sqrt3)^4-1}\right)^{1/4}G\\ \quad\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) &= 2^{1/4}\left(\Big(\tfrac{1+\sqrt5}2\Big)^{12}-\sqrt{\Big(\tfrac{1+\sqrt5}2\Big)^{24}-1}\right)^{1/4}G\end{align}$$ and so on, with fundamental units \(U_3 = 2+\sqrt3\) and \(U_5= \frac{1+\sqrt5}2\), and where the \(\lambda(\tau)\) are actually radicals. Curiously, \(\lambda(i)\ = \lambda\big(\sqrt{3+4i}\big) = \frac12\).